Mathematics for the interested outsider

Subspaces from Irreducible Representations

Because of Maschke’s theorem we know that every representation of a finite group can be decomposed into chunks that correspond to irreducible representations:

where the are pairwise-inequivalent irreps. But our consequences of orthonormality prove that there can be only finitely many such inequivalent irreps. So we may as well say that is the number of them and let a multiplicity be zero if doesn’t show up in at all.

Now there’s one part of this setup that’s a little less than satisfying. For now, let’s say that is an irrep itself, and let be a natural number for its multiplicity. We’ve been considering the representation

made up of the direct sum of copies of . But this leaves some impression that these copies of actually exist in some sense inside the sum. In fact, though inequivalent irreps stay distinct, equivalent ones lose their separate identities in the sum. Indeed, we’ve seen that

That is, we can find a copy of lying “across” all copies in the sum in all sorts of different ways. The identified copies are like the basis vectors in an -dimensional vector space — they hardly account for all the vectors in the space.

We need a more satisfactory way of describing this space. And it turns out that we have one:

Here, the tensor product is over the base field , so the “extra action” by on makes this into a -module as well.

This actually makes sense, because as we pass from representations to their characters, we also pass from “plain” vector spaces to their dimensions, and from tensor products to regular products. Thus at the level of characters this says that adding copies of an irreducible character together gives the same result as multiplying it by , which is obviously true. So since the two sides have the same characters, they contain the same number of copies of the same irreps, and so they are isomorphic as asserted.

Actually, any vector space of dimension will do in the place of here. And we have one immediately at hand: itself. That is, if is an irreducible representation then we have an isomorphism:

As an example, if is any representation and is any irrep, then we find

We can reassemble these subspaces to find

Notice that this extends our analogy between spaces and inner products. Indeed, if we have an orthonormal basis of a vector space of dimension , we can decompose any vector as

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